# Page:LorentzGravitation1916.djvu/16

draw a line perpendicular to ${\displaystyle \sigma _{2}}$ and ${\displaystyle \sigma _{1}}$. Let ${\displaystyle B_{1}}$ be the point, where it cuts thus last, plane, the "base", and ${\displaystyle A_{1}}$ the point where this plane is encountered by the generating line through ${\displaystyle A_{2}}$. If then ${\displaystyle \angle A_{1}A_{2}B_{1}=\vartheta }$, we have

 ${\displaystyle {\overline {A_{2}B_{1}}}={\overline {A_{2}A_{1}}}\cos \vartheta }$ (13)

The strokes over the letters indicate the absolute values of the distances ${\displaystyle A_{2}B_{1}}$ and ${\displaystyle A_{2}A_{1}}$.

It can be shown (§ 8) that, all quantities being expressed in natural units, the "volume" of the prism ${\displaystyle P}$ is found by taking the product of the numerical values of the base ${\displaystyle \sigma _{1}}$ and the "height" ${\displaystyle A_{2}B_{1}}$.

Let now linear three-dimensional extensions perpendicular to ${\displaystyle A_{1}A_{2}}$ be made to pass through ${\displaystyle A_{1}}$ and ${\displaystyle A_{2}}$. From these extensions the lateral boundary of the prism cuts the parts ${\displaystyle \sigma '_{1}}$ and ${\displaystyle \sigma '_{2}}$ and these parts, together with the lateral surface, enclose a new prism ${\displaystyle P'}$, the volume of which is equal to that of ${\displaystyle P}$. As now the volume of ${\displaystyle P'}$ is given by the product of ${\displaystyle {\overline {A_{2}A_{1}}}}$ and ${\displaystyle \sigma '_{1}}$, we have with regard to (13)

${\displaystyle \sigma '_{1}=\sigma {}_{1}\cos \vartheta }$

If now we remember that, if a vector perpendicular to ${\displaystyle \sigma _{1}}$ is projected on the generating line, the ratio between the projection and the vector itself (viz. between their absolute values) is given by ${\displaystyle \cos \vartheta }$ and that a connexion similar to that which was found above between a normal section ${\displaystyle \sigma '_{1}}$ of the prism and ${\displaystyle \sigma _{1}}$, also exists between ${\displaystyle \sigma '_{1}}$ and any other oblique section, we easily find the following theorem:

Let ${\displaystyle \sigma }$ and ${\displaystyle {\bar {\sigma }}}$ be two arbitrarily chosen linear three-dimensional sections of the prism, ${\displaystyle \mathrm {N} }$ and ${\displaystyle {\bar {\mathrm {N} }}}$ two vectors, perpendicular to ${\displaystyle \sigma }$ and ${\displaystyle {\bar {\sigma }}}$ resp. and of the same length, ${\displaystyle S}$ and ${\displaystyle {\bar {S}}}$ the absolute values of the projections of ${\displaystyle \mathrm {N} }$ and ${\displaystyle {\bar {\mathrm {N} }}}$ on a generating line. Then we have

 ${\displaystyle S\sigma ={\bar {S}}{\bar {\sigma }}}$ (14)

§ 19. After these preliminaries we can show that the left hand side of (10) is equal to 0, if the numbers ${\displaystyle g_{ab}}$ are constants and if moreover both the rotation ${\displaystyle \mathrm {R} _{e}}$ and the rotation ${\displaystyle \mathrm {R} _{h}}$ are everywhere the same. For the two parts of the integral the proof may be given in the same way, so that it suffices to consider the expression

 ${\displaystyle \int \left[\mathrm {R} _{e}\cdot \mathrm {N} \right]_{x}d\sigma }$ (15)

Let ${\displaystyle X_{1},\dots X_{4}}$ be the components of the vector ${\displaystyle \mathrm {N} }$, expressed in ${\displaystyle x}$-units. From the distributive property of the vector product it then follows that each of the four components of